Fix 2004/kopczynski and format fix README.md
The fix is that we cannot pass -ON to the compiler. With -ON it enters an infinite loop as the variable l goes up, not down. Format fix README.md. Add kopczynski-a to show what happens when it's not a number.
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@@ -66,7 +66,7 @@ CINCLUDE=
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# Optimization
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#
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OPT= -O3
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OPT=
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# Default flags for ANSI C compilation
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#
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+76
-71
@@ -1,12 +1,8 @@
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# Best one-liner
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Eryk Kopczynski
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Faculty of Mathematics, Informatics and Mechanics
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ul. Banacha 2
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02-097 Warszawa
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Poland
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http://www.mimuw.edu.pl/~erykk/
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Eryk Kopczynski
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Poland
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<http://www.mimuw.edu.pl/~erykk/>
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## To build:
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@@ -20,6 +16,11 @@ make
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./kopczynski < input_file
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```
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[Cody Boone Ferguson](/winners.html#Cody_Boone_Ferguson) reported that this
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entry cannot be optimised by the compiler if it's to work. He also added the
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file [kopczynski-a](kopczynski-a) to demonstrate what happens when art more like
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a letter is fed to the program. Thank you Cody!
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## Try:
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```sh
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@@ -31,6 +32,14 @@ make
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./kopczynski < kopczynski-11
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```
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Also try:
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```sh
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./kopczynski < kopczynski-a
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```
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What happens?
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### Alternate code
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The judges modified the program to print its result. The original
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@@ -47,7 +56,7 @@ Then try:
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echo $?
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```
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## Judges' remarks
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## Judges' remarks:
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What is in a line? A lot when you obfuscate the way Eryk Kopczynski
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did it. This small one line program is an outstanding technical
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@@ -69,10 +78,10 @@ Here is a related puzzle: how to continue the sequence:
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1, 0, 0, 0, 0, 0, 1, 0, 2, 1, ...
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With certain arguments, you can alter the exit value. Can you figure
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With certain arguments, you can alter the exit value / output. Can you figure
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out how and why?
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## Author's remarks
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## Author's remarks:
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The program proves it is not as hard to recognize numbers. The current
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one liner version should correctly recognize all numbers from 8 to 11.
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@@ -81,112 +90,108 @@ recognize negatives, zeros which have dots or slashes inside them,
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digits should be separated, and numbers should be "complete".)
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This program returns the number as its exit status. (In most shells,
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you can see the exit status of the last command with 'echo $?' (or
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just '$?').)
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you can see the exit status of the last command with `echo $?` (or
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just `$?`).)
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[[NOTE: The judges altered the primary version to output a number.]]
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**NOTE: The judges altered the primary version to output a number.**
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On the input, give numbers formed as a sequences of empty spaces and #
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signs (you can also use dots instead of spaces), ending with an empty
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line. (No number has an empty line in it.) The program should return
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the number from 8 to 11 that it thinks it is.
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* The line is 131 bytes long (in 1986 165 bytes was OK, and the size
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* The line is 131 bytes long (in 1986, 165 bytes was OK, and the size
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limit has grown from 1986...).
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* I have decided that it is better for this program to return the
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number, not to print it, but if you think otherwise, the change would
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cost 6 chars (plus a warning). Change 'return *' to 'printf("%d",*)'.
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('$?' is shorter)
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cost 6 chars (plus a warning). Change `return *` to `printf("%d",*)`.
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(`$?` is shorter)
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* I achieved some additional tightness by '-Dif=while', saved 9 chars
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of source for cost of 11 chars of build file. If I used '-DB=while',
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I would save 12 for 10, but I think the current version is slightly
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more obfuscated.
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* I achieved some additional tightness by `-Dif=while` which saved 9 chars of
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source for the cost of 11 chars of the [Makefile](Makefile). If I used
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`-DB=while`, I would save 12 for 10, but I think the current version is slightly
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more obfuscated.
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* A remark on portability:
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### Portability
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It only requires two's compliment arithmetics, as far as I
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know. The maximum length of an input line is N-9, where
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N is the size of int in bits (you can try to compile with
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an option '-Dint=long long' for longer lines).
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It only requires two's complement arithmetics, as far as I know. The maximum
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length of an input line is `N-9`, where `N` is the size of `int` in bits
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(you can try to compile with an option `'-Dint=long long'` for longer
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lines).
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-- Algorithm & Obfuscation --
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### Algorithm & Obfuscation
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1. This program works by calculating the Euler characteristics of input,
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plus 9. It is the number of digits (i.e. separated #-regions) minus the
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number of holes in them. It turns out that this algorithm works
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correctly for all numbers from 8 to 11.
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1. This program works by calculating the [Euler
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characteristics](https://en.wikipedia.org/wiki/Euler_characteristic) of input,
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plus 9. It is the number of digits (i.e. separated #-regions) minus the number
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of holes in them. It turns out that this algorithm works correctly for all
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numbers from 8 to 11.
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2. We know that Euler characteristics is the number of "vertexes"
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2. We know that [Euler
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characteristics](https://en.wikipedia.org/wiki/Euler_characteristic) is the number of "vertexes"
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#
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#
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minus the number of "edges"
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## # #. .#
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# .# #. ('.' must be empty)
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## # #. .#
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# .# #. ('.' must be empty)
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plus the number of "faces"
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##
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##
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##
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##
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3. If we fill all these patterns to 2x2 squares,
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#? ## #? #. .# ##
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?? ?? #? .# #. ##
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+1 -1 -1 -1 -1 +1
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#? ## #? #. .# ##
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?? ?? #? .# #. ##
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+1 -1 -1 -1 -1 +1
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substitute all possible "values" of '?' and sum up, we get that it is
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substitute all possible "values" of `?` and sum them up, we get that it is
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enough to calculate:
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#. .# ##
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.. #. #.
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+1 -1 -1
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#. .# ##
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.. #. #.
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+1 -1 -1
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Note now it is simple and obfuscated enough to be used as
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a "magic trick".
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4. I make a 16-character array; when indexed by 4-bit value which
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represents the 2x2 squares, it will return a number to add/subtract
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{-1,0,+1}
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`{-1,0,+1}`
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5. We convert lines of input to bitmaps (N-bit integers). Additionally,
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it is heavily obfuscated by using negative numbers (two's compliment).
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For example, it turns out that 6 in "z=6" in the only 1-digit number
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which works here (and it's not only because we have "z>5" in the
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following while). The conversion routine also sets n=8 in an obfuscated
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and concise way.
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5. We convert lines of input to bitmaps (`N-bit` integers). Additionally, it is
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heavily obfuscated by using negative numbers (two's complement). For example,
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it turns out that `6` in `z=6` is the only 1-digit number which works here (and
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it's not only because we have `z>5` in the following `while`). The conversion
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routine also sets `n=8` in an obfuscated and concise way.
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5. Since we always return the number from 8 to 11, we can calculate
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modulo 4. Instead of 16 {-1,0,+1} values, we use a 15-character string
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which agrees modulo 4 to the array in (4). (NUL fits very well as the
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16th character, so we use 15-character string.)
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6. Since we always return the number from 8 to 11, we can calculate modulo 4.
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Instead of `16 {-1,0,+1}` values, we use a 15-character string which agrees
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modulo 4 to the array in (`4`). (`NUL` fits very well as the 16th character, so
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we use 15-character string.)
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6. Since the rule found in (3) did not produce any funny messages, I
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changed it until it did. (The # character to find vertexes is in the
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top right corner, not the top left one.) The following messages will
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also work instead of "has dirtiest IF":
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7. Since the rule found in (3) did not produce any funny messages, I changed it until it did. (The `#` character to find vertexes is in the top right corner, not the top left one.) The following messages will also work instead of `has dirtiest IF`:
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HighlandYachter
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TestDirtiestTen
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has lintiest IF
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TighterLaughter
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HustlerDaughter
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TighterDemoThen
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HighlandYachter
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TestDirtiestTen
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has lintiest IF
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TighterLaughter
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HustlerDaughter
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TighterDemoThen
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They can be mixed, of course!
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7. [[NOTE: Item 7 was removed by the judges so as to not spoil the
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challenge of figuring how arguments can be used to alter
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the exit value.]]
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8. NOTE: item 8 was removed by the judges so as to not spoil the challenge of
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figuring how arguments can be used to alter the exit value.
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8. The variable names are confusing.
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9. The variable names are confusing.
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9. Finally, the program tries to be as concise as possible. Some
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obfuscation comes from this. For additional obfuscation and
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conciseness, 'if' means 'while'.
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10. Finally, the program tries to be as concise as possible. Some obfuscation
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comes from this. For additional obfuscation and conciseness, `if` means `while`.
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## Copyright and CC BY-SA 4.0 License:
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@@ -0,0 +1,8 @@
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####
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######
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## ##
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### ###
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##########
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## ##
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## ##
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